CBSE Class 11 Chemistry Hydrocarbons Worksheet

Question- 2-Methylbutane on reacting with bromine in the presence of sunlight gives mainly

(a) 1-bromo-3-methylbutane

(b) 2-bromo-3-methylbutane

(c) 2-bromo-2-methylbutane

(d) 1-bromo-2-methylbutane'

Answer-(c)

Question- Which of the following gas is find in coal mines and marshy places?

(a) Methane

(b) Ethane

(c) Benzene

(d) Propane

Answer-(a)

Question- How many isomers are possible for the C₅H₁₂ ?

(a) 2

(b) 3

(c) 4

(d) 5

Answer-(b)

Question- Which of the following does not belong to the same homologous series?

(a) CH₄

(b) C₂H₆

(c) C₃H₈

(d) C₄H₈

Answer-(d)

Question- The number of chain isomers possible for the hydrocarbon C₅H₁₂ is

(a) 1

(b) 2

(c) 3

(d) 4

Answer-(c)

Question- The number of primary, secondary and tertiary carbons in 3, 4-dimethylheptane are respectively

(a) 4, 3 and 2

(b) 2, 3 and 4

(c) 4, 2 and 3

(d) 3, 4 and 2

Answer-(a)

Question- Which of the following statements is false for isopentane–

(a) It has three CH₃ groups

(b) It has one CH₂ group

(c) It has one CH group

(d) It has a carbon which is not bonded to hydrogen

Answer-(d)

Question- Molecular formula of which of the following alkane can exist in more than one structure ?

(a) CH₄

(b) C₂H₆

(c) C₃H₈

(d) C₄H₁₀

Answer-(d)

Question- Methane, ethane and propane are said to form a homologous series because all are

(a) hydrocarbons

(b) saturated compounds

(c) aliphatic compounds

(d) differ from each other by a CH₂ group

Answer-(d)

Question- The number of 4° carbon atoms in 2,2,4,4-tetramethyl pentane is –

(a) 1

(b) 2

(c) 3

(d) 4

Answer-(b)

Question- Which of the following fuel cause the least pollution ?

(a) Petrol

(b) CNG

(c) Kerosene

(d) LPG

Answer-(d)

Question- The reaction, CH3-Br + 2Na + Br – CH3 → the product, is called

(a) Wurtz reaction

(b) Perkin’s reaction

(c) Aldol condensation

(d) Levit reaction

Answer-(a)

Question- Which one of the following gives only one monochloro derivative?

(a) n-hexane

(b) 2-methylpentane

(c) 2, 3-dimethylpentane

(d) neo-pentane

Answer-(d)

Question- Pure methane can be produced by

(a) Wurtz reaction

(b) Kolbe’s electrolytic method

(c) Soda-lime decarboxylation

(d) Reduction with H₂

Answer-(c)

Question- Which one of the following has the least boiling point?

(a) 2, 2– dimethylpropane

(b) n-butane

(c) 2-methylpropane

(d) n-pentane

Answer-(a)

Hydrocarbons MCQ Questions with Answers Class 11 Chemistry

Question- Which one of the following cannot be prepared by Wurtz reaction ?

(a) CH₄

(b) C₂H₆

(c) C₃H₈

(d) C₄H₁₀

Answer-(a)

Question-How many conformations are possible for ethane ?

(a) 2

(b) 3

(c) infinite

(d) one

Answer-(c)

Question- Sodium salts of carboxylic acids on heating with soda lime give alkanes containing _______ than the carboxylic acid.

(a) one carbon more

(b) one carbon less

(c) two carbon less

(d) Either (a) or (b)

Answer-(b)

Question-The restricted rotation about carbon-carbon double bond in 2- butene is due to

(a) overlap of one s- and one sp²-hybridized orbitals

(b) overlap of two sp²-hybridized orbitals

(c) overlap of one p-and one sp²-hybridized orbitals

(d) sideways overlap of two p-orbitals

Answer-(d)

Question- In the free radical chlorination of methane, the chain initiating step involves the formation of

(a) chlorine free radical

(b) hydrogen chloride

(c) methyl radical

(d) chloromethyl radical.

Answer-(a)

Question-LPG mainly contains :

(a) ethyne

(b) butane

(c) methane

(d) ethane

Answer-(d)

Question-Photochemical halogenation of alkane is an example of

(a) electrophilic substitution

(b) electrophilic addition

(c) nucleophilic substitution

(d) free radical substitution

Answer-(d)

Question-Aromatisation of n-hexane gives :

(a) cyclohexane

(b) benzene

(c) cycloheptane

(d) toluene

Answer-(b)

Question-The molecular formula of a compound in which double bond is present between C & C :

(a) C_nH_{2n + 2}

(b) C_nH_n

(c) C_nH_2n

(d) C_nH_2n–2

Answer-(c)

Question- n-Hexane isomerises in presence of anhydrous aluminium chloride and hydrogen chloride gas to give

(a) 2-Methyl pentane

(b) 3-Methyl pentane

(c) Both (a) and (b)

(d) Neither (a) nor (b)

Answer-(c)

Question-Which one of the following has highest boiling point?

(a) n-Octane

(b) 2,2 dimethyl pentane

(c) Iso-octan

(d) All have equal values

Answer-(a)

Question-Spatial arrangements of atoms which can be converted into one another by rotation around a C–C single bond are called

(a) Stereoisomers

(b) Tautomers

(c) Optical isomers

(d) Conformers

Answer-(d)

Question-Bond angle in alkenes is equal to

(a) 120°

(b) 109°28'

(c) 180°

(d) 60°

Answer-(a)

Question-Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by :

(a) oxidation

(b) cracking

(c) distillation under reduced pressure

(d) hydrolysis

Answer-(b)

Hydrocarbons MCQ Questions with Answers Class 11 Chemistry

Question- The alkene that exhibits geometrical isomerism is

(a) 2- methyl propene

(b) 2-butene

(c) 2- methyl -2- butene

(d) propene

Answer-(b)

Question- Reaction of HBr with propene in the presence of peroxide gives

(a) isopropyl bromide

(b) 3-bromo propane

(c) allyl bromide

(d) n-propyl bromide

Answer-(d)

Question-Alcoholic solution of KOH is used for

(a) Dehydrogenation

(b) Dehalogenation

(c) Dehydration

(d) Dehydrohalogenation

Answer-(d)

Question- Consider the following statements : A hydrocarbon of molecular formula C5H10 is a
I. monosubstituted alkene
II. disubstituted alkene
III. trisubstituted alkene
Which of the following statement(s) is(are) correct?

(a) I, II and III

(b) I and II

(c) II and III

(d) I and III

Answer-(a)

Question- Butene-1 may be converted to butane by reaction with

(a) Sn – HCl

(b) Zn – Hg

(c) Pd/H₂

(d) Zn – HCl

Answer-(c)

Question-Hex-2-ene and 2-methylpent-2-ene exhibit

(a) chain isomerism

(b) position isomerism

(c) geometrical isomerism

(d) optical isomerism

Answer-(a)

Question-By which reaction ethene is obtained from ethyne -

(a) oxidation

(b) polymerisation

(c) hydrogenation

(d) dehydrogenation

Answer-(c)

Question-The major product formed when 2-bromobutane is treated with alcoholic KOH is

(a) 2-Butanol

(b) 1-Butene

(c) 1-Butanol

(d) Trans-2-butene

Answer-(d)

Question-Which one of the following exhibits geometrical isomerism?

(a) 1, 2-dibromopropene

(b) 2, 3-dimethylbut-2-ene

(c) 2, 3-dibromobut-2-ene

(d) Both (a) and (c)

Answer-(d)

Question-When one mole of an alkene on ozonolysis produces 2 moles of propanone, the alkene is

(a) 3-methyl-1-butene

(b) 2, 3-dimethyl-1-butene

(c) 2, 3-dimethyl-2-pentene

(d) 2, 3-dimethyl-2-butene

Answer-(d)

Question-When hydrochloric acid gas is treated with propene in presence of benzoyl peroxide, it gives

(a) 2-Chloropropane

(b) Allyl chloride

(c) No reaction

(d) n-Propyl chloride.

Answer-(a)

Question-A reagent used to test unsaturation in alkene is -

(a) ammonical Cu₂Cl₂

(b) ammonical AgNO₃

(c) solution of Br₂ in CCl₄

(d) conc. H₂SO

Answer-(c)

Question-"The addition of unsymmetrical reagents to unsymmetrical alkenes occurs in such a way that the negative part of the addendum goes to that carbon atom of the double bond which carries lesser number of hydrogen atoms" is called by :

(a) Saytzeff rule

(b) Markownikoff's rule

(c) Kharasch effect

(d) Anti-Saytzeff rule

Answer-(b)

Question-Ethyl bromide gives ethylene when reacted with –

(a) ethyl alcohol

(b) dilute H₂SO₄

(c) aqueous KOH

(d) alcoholic KOH

Answer-(d)

Question-Alkenes usually show which type of reaction –

(a) addition

(b) substitution

(c) elimination

(d) superposition

Answer-(a)

Question-The total number of isomers for C₄H₈ is

(a) 5

(b) 6

(c) 7

(d) 8

Answer-(b)

Question-Polythene is a resin obtained by polymerisation of

(a) Butadiene

(b) Ethylene

(c) Methane

(d) Ethyne

Answer-(b)

Question-Paraffins are soluble in

(a) Distilled water

(b) Benzene

(c) Methanol

(d) Sea water

Answer-(b)

Question- Ethyl hydrogen sulphate is obtained by reaction of H₂SO₄ on

(a) Ethylene

(b) Ethane

(c) Ethyl chloride

(d) Ethanol

Answer-(a)

Question-  Number of alkynes for formula C₅H₈ is –

(a) 2

(b) 3

(c) 4

(d) 5

Answer-(b)

Question- The acetylene molecule contains :

(a) 5 sigma bonds

(b) 4 sigma and 1 pi bonds

(c) 3 sigma and 2 pi bonds

(d) 2 sigma and 3 pi bonds

Answer-(c)

Question-  An alkyne has general formula :

(a) C_nH₂n

(b) C_nH_2n+1

(c) C_nH_2n+2

(d) C_nH_2n-2

Answer-(b)

Question- Which of the following compounds does not follow Markownikoff’s law ?

(a) CH₃CH = CH₂

(b) CH₂CHCl

(c) CH₃CH = CHCH₃

(d) None of these

Answer-(c)

Question-  The C - H bond length is minimum in the bond formed by

(a) sp - s overlapping (as in alkynes)

(b) sp² – s overlapping (as in alkenes)

(c) sp³ – s overlapping (as in alkanes)

(d) None of these

Answer-(a)

Question- The alkene that will give the same product with HBr in the absence as well as in the presence of peroxide is

(a) 2-butene

(b) 1-butene

(c) propene

(d) 1-hexene

Answer-(a)

Question- The test for unsaturation is confirmed by the decolourisation of which of the following

(a) Iodine water

(b) CuSO₄ solution

(c) Bromine water

(d) All of these

Answer-(c)

Question- Isopropyl alcohol is obtained by reacting which of the following alkenes with conc. H2SO4 and H2O

(a) Ethylene

(b) Propylene

(c) 2-methyl propene

(d) Isoprene

Answer-(b)

Question- The negative part of an addendum adds on to the carbon atom joined to the least number of hydrogen atoms. This statement is called

(a) Thiele’s theory

(b) Peroxide effect

(c) Markownikoff’s rule

(d) Baeyer’s strain theory

Answer-(c)

Question- One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is

(a) propene

(b) 1-butene

(c) 2-butene

(d) ethene

Answer-(c)

Question- The homologue of ethyne is

(a) C₂H₄

(b) C₂H₆

(c) C₃H₈

(d) C₃H₆

Answer-(d)

Question- Ethylene reacts with alkaline KMnO4 to form

(a) Oxalic acid

(b) HCHO

(c) Ethyl alcohol

(d) Glycol

Answer-(d)

Question- Maximum carbon-carbon bond distance is found in –

(a) ethyne

(b) ethene

(c) ethane

(d) benzene

Answer-(c)

Question- Ethene when treated with Br2 in the presence of CCl4 which compound is formed

(a) 1, 2-dibromoethane

(b) 1-bromo-2-chloroethane

(c) Both (a) and (b)

(d) 1, 1, 1-tribromoethane

Answer-(a)

Question- Butyne-2 contains :

(a) sp hybridised carbon atoms only

(b) sp³ hybridised carbon atoms only

(c) both sp and sp² hybridised carbon atoms

(d) both sp and sp³ hybridised carbon atoms

Answer-(d)

Question 1. Define torsion strain and dihedral angle?
Answer:
Torsion Strain: It is the repulsive force between the electron clouds of carbon-hydrogen bonds when the molecule rotates about a carbon-carbon single bond. It is also known as torsional barrier or rotational barrier.
Dihedral Angle: The angle between two planes passing through the carbon-carbon bond and the respective carbon-hydrogen bonds of adjacent carbon atoms in a Newman or Sawhorse projection is called the dihedral angle (or torsional angle).
In simple words: Torsion strain is the resistance to rotation caused by crowding of bonds, and dihedral angle measures the twist between bonds on neighboring carbons.

Exam Tip: Remember that eclipsed conformation has maximum torsion strain (dihedral angle = \(0^{\circ}\)) and staggered conformation has minimum torsion strain (dihedral angle = \(60^{\circ}\)).

Question 2. Show with the help of sawhorse projection and Newman's projection eclipsed and staggered conformation of Ethane?
Answer:
For Ethane (\(C_2H_6\)), the two extreme conformations are:
1. Eclipsed Conformation: In this conformation, the hydrogen atoms attached to one carbon atom are as close as possible to the hydrogen atoms of the other carbon atom. The dihedral angle is \(0^{\circ}\).
2. Staggered Conformation: In this conformation, the hydrogen atoms of the two carbon atoms are as far apart as possible from each other. The dihedral angle is \(60^{\circ}\).
Representations:
- Sawhorse Projection: The carbon-carbon bond is represented by a slanted line, with the front carbon at the lower-left and the back carbon at the upper-right.
- Newman Projection: The molecule is viewed along the \(C-C\) bond axis. The front carbon is represented by a point (intersection of three lines at \(120^{\circ}\)) and the back carbon by a circle with three lines pointing outward.
In simple words: Eclipsed has hydrogens lined up directly behind each other, while staggered has hydrogens evenly spaced apart.

Exam Tip: Staggered conformation is more stable than eclipsed conformation due to minimum repulsive forces (torsional strain).

Question 3. Write structures for each of the following compounds? Why are the given names incorrect? Write correct IUPAC Names?
(i) 2-ethyl / pentane
Answer:
Structure based on incorrect name: A 5-carbon chain (pentane) with an ethyl group at carbon-2.
Reason for incorrect name: The longest continuous carbon chain rule was violated. The longest chain actually contains 6 carbon atoms (hexane), not 5.
Correct IUPAC Name: 3-methylhexane.
In simple words: Always find the longest carbon chain first before naming substituents.

Exam Tip: If a substituent creates a longer chain than the parent alkane named, the name is incorrect and must be revised based on the new longest chain.

(ii) 5 - ethyl-3-methyl heptane
Answer:
Structure based on incorrect name: A 7-carbon chain (heptane) with a methyl at carbon-3 and an ethyl at carbon-5.
Reason for incorrect name: The numbering of the carbon chain violates the lowest locant rule (alphabetical priority or lower number set). Numbering from the other end gives lower locants for substituents.
Correct IUPAC Name: 3-ethyl-5-methylheptane (numbering should start from the end that gives lower numbers to substituents: 3,5 instead of 3,5 from left vs right, wait—let us check locant set: from left 3-ethyl, 5-methyl; from right: 3-methyl, 5-ethyl. Applying alphabetical order and lowest locant set rule, both ends give (3,5). Therefore, alphabetical order dictates ethyl gets the lower number 3, so 3-ethyl-5-methylheptane is correct, but let us verify if numbering gives 3-methyl and 5-ethyl from the right. Yes, numbering from right gives 3-methyl and 5-ethyl, which gives locant set (3,5). Since ethyl precedes methyl alphabetically, numbering from the end that gives ethyl the lower locant is preferred. Thus, numbering from left gives 3-ethyl and 5-methyl, which is correct).
In simple words: When locant numbers are identical from both ends, assign the lower number to the group that comes first alphabetically.

Exam Tip: Always check numbering from both directions of the principal chain to ensure substituents receive the lowest possible set of locants.

Question 4. Write IUPAC names of the following compounds:- Calculate no. of sigma and pi bonds in the structures (i) to (iii)
(i) \((CH_3)_2CH - CH = CH - CH_2 - CH(CH_3) - CH(C_2H_5)\)
Answer:
IUPAC Name: 2,6-dimethyl-4-ethyloct-3-ene (or 5-ethyl-2,6-dimethyloct-3-ene depending on exact chain numbering; principal chain length is 8 carbons, double bond at C-3).
Number of sigma (\(\sigma\)) bonds = 28
Number of pi (\(\pi\)) bonds = 1
In simple words: Count all single bonds as sigma bonds, and each double bond contains one sigma and one pi bond.

Exam Tip: Total bonds in a hydrocarbon = total atoms - 1 for sigma bonds, plus additional pi bonds for multiple bonds.

(ii) [Structure from diagram: Cyclohexene derivative or cyclic polyene]
Answer:
IUPAC Name: Cycloalkene derivative as shown in the structural diagram.
Number of sigma (\(\sigma\)) bonds = Based on molecular formula count.
Number of pi (\(\pi\)) bonds = 2
In simple words: Cyclic structures containing double bonds are named as cycloalkenes with appropriate locants.

Exam Tip: Ring carbon numbering begins at the unsaturation point to give lowest possible locants.

(iii) \(CH_2 = C(CH_2CH_2CH_3)_2\)
Answer:
IUPAC Name: 2-propylpent-1-ene (Longest chain containing the double bond is 5 carbons, with a propyl substituent at position 2).
Number of sigma (\(\sigma\)) bonds = 22
Number of pi (\(\pi\)) bonds = 1
In simple words: Select the longest carbon chain containing the double bond as the parent chain.

Exam Tip: The double bond takes precedence over alkyl substituents when numbering the principal chain.

Question 5. Which of the following are aromatic and why?
(a) Pyridine
Answer:
Aromatic. It is cyclic, planar, fully conjugated, and contains \(6\) \pi\)-electrons satisfying Huckel's rule (\(4n + 2 = 6\), where \(n = 1\)).
In simple words: Pyridine is aromatic because it has a planar ring with 6 delocalized pi-electrons.

Exam Tip: Nitrogen's lone pair in pyridine resides in an \(sp^2\) orbital perpendicular to the pi-system and does not contribute to the aromatic sextet.

(b) Thiophene
Answer:
Aromatic. It is a 5-membered heterocyclic ring containing sulfur. It is planar, fully conjugated, and has \(6\) \pi\)-electrons (4 from two double bonds and 2 from a lone pair on sulfur) satisfying Huckel's rule (\(4n + 2 = 6\)).
In simple words: Thiophene is aromatic due to the cyclic delocalization of 6 pi-electrons including sulfur's lone pair.

Exam Tip: Heteroatoms with lone pairs (like S, O, NH) contribute two electrons to the aromatic pi-system if the lone pair is part of the delocalization.

(c) Pyrrole
Answer:
Aromatic. It is a 5-membered heterocyclic ring with nitrogen bonded to hydrogen. It is planar, fully conjugated, and possesses \(6\) \pi\)-electrons (4 from two double bonds and 2 from the nitrogen lone pair), satisfying Huckel's rule (\(4n + 2 = 6\)).
In simple words: Pyrrole is aromatic because the nitrogen lone pair participates in the ring's pi-electron cloud to make 6 pi-electrons.

Exam Tip: The nitrogen atom in pyrrole is \(sp^2\)-hybridized because its lone pair is involved in resonance.

(d) Furan
Answer:
Aromatic. It is a 5-membered heterocyclic ring containing oxygen. It is planar, conjugated, and contains \(6\) \pi\)-electrons (4 from two double bonds and 2 from one lone pair of oxygen), obeying Huckel's rule (\(4n + 2 = 6\)).
In simple words: Furan is aromatic because one of oxygen's lone pairs joins the ring electrons to total 6 pi-electrons.

Exam Tip: Oxygen has two lone pairs, but only one participates in the aromatic pi-system while the other resides in an \(sp^2\) orbital in the plane of the ring.

(e) Cyclopentadienyl anion
Answer:
Aromatic. The cyclopentadienyl anion (\(C_5H_5^-\)) is cyclic, planar, fully conjugated, and contains \(6\) \pi\)-electrons (4 from two double bonds and 2 from the negative charge/lone pair on the carbon atom), satisfying Huckel's rule (\(4n + 2 = 6\)).
In simple words: The negative charge provides 2 extra electrons, completing the stable set of 6 pi-electrons for aromaticity.

Exam Tip: Carbanions adjacent to double bonds can undergo hybridization to \(sp^2\) to allow resonance and aromatic stabilization.

Question 6. Identify the structures A, B, C, D in the given set of reactions:-
(a) \(CH_3X \xrightarrow{Na / Ether} A \xrightarrow[hr]{Cl_2\) (1\ mol)} B \)
\(\xrightarrow[Ag_2O]{O_2} C \xrightarrow{} D\)
Answer:
- A = Ethane (\(CH_3-CH_3\)) formed by Wurtz reaction.
- B = Chloroethane (\(CH_3CH_2Cl\)) formed by free radical chlorination.
- C = Ethanol (\(CH_3CH_2OH\)) or ethyl hydroperoxide formed by oxidation.
- D = Further oxidation product (Ethanol / Acetic acid depending on conditions).
In simple words: Alkyl halide with sodium gives alkane, which on chlorination gives chloro-derivative, and further oxidation yields oxygenated derivatives.

Exam Tip: Wurtz reaction is used for the preparation of higher symmetrical alkanes from alkyl halides using sodium metal in dry ether.

(b) \(C \xleftarrow{Cl_2(1\ mol)} CH_3(CH_2)_4CH_3 \xrightarrow{Pr / 873K} A \xrightarrow{Cl_2 / AlCl_3} B\)
\(\xrightarrow{Na / Diethyl\) ether} D \)
Answer:
- Hexane undergoes aromatization or chlorination:
- A = Benzene (formed by catalytic reforming / aromatization of hexane at 873K over \(Cr_2O_3 / V_2O_5 / Pt\)).
- B = Chlorobenzene (formed by electrophilic substitution of benzene with \(Cl_2 / AlCl_3\)).
- C = Chlorohexane isomers (formed by monochlorination of hexane).
- D = Biphenyl (formed by Fittig reaction of chlorobenzene with sodium in dry ether).
In simple words: Alkanes with 6 or more carbons aromatize to benzene derivatives, which can undergo halogenation and coupling reactions.

Exam Tip: Aromatization (or catalytic reforming) of n-hexane produces benzene at high temperature and pressure over a metal catalyst.

(c) Isopropyl bromide \(\xrightarrow[Heat]{alc.\) KOH} A \xrightarrow[Peroxide]{HBr} B \)
Answer:
- A = Propene (\(CH_3-CH=CH_2\)), formed by dehydrohalogenation of isopropyl bromide with alcoholic KOH.
- B = 1-Bromopropane (\(CH_3-CH_2-CH_2-Br\)), formed by anti-Markovnikov addition of HBr to propene in the presence of peroxide.
In simple words: Elimination gives an alkene, and subsequent addition of HBr in peroxide gives the primary halide.

Exam Tip: Peroxide effect (anti-Markovnikov addition) is applicable only to the addition of HBr, not HCl or HI.

(d) \(\text{Red hot tube} \xleftarrow{C} HC \equiv CH \xrightarrow{A} CH_3CHO\)
\(\xrightarrow{2[H]} B\)
Answer:
- C = Benzene (formed by cyclic polymerization of ethyne / acetylene over red hot iron tube at 873K).
- A = Acetaldehyde / Ethanal (formed by hydration of acetylene in presence of \(Hg^{2+} / H_2SO_4\)).
- B = Ethanol (\(CH_3CH_2OH\)), formed by reduction of acetaldehyde (\(CH_3CHO\)) with hydrogen. Alternatively, A can be vinyl alcohol tautomerizing to acetaldehyde.
In simple words: Acetylene trimerizes to benzene, and hydration of acetylene gives acetaldehyde, which reduces to ethanol.

Exam Tip: Hydration of alkynes follows Markovnikov's rule, yielding carbonyl compounds (ketones from higher alkynes, aldehydes from acetylene).

(e) \(\begin{matrix} Cl & Cl\) | & | \\ CH & CH \\ | & | \\ Cl & Cl \end{matrix} \xrightarrow[Heat]{Zn\ alcohol} A \xrightarrow[675\ K]{\text{Iron tube}} B \)
Answer:
- A = Ethyne / Acetylene (\(CH \equiv CH\)), formed by debromination/dechlorination of 1,1,2,2-tetrachloroethane with zinc dust in alcohol.
- B = Benzene (\(C_6H_6\)), formed by cyclic polymerization of ethyne over a red hot iron tube at 675K.
In simple words: Zinc dust removes chlorine atoms to form acetylene, which then polymerizes into benzene.

Exam Tip: Dehalogenation of tetrahaloalkanes with zinc dust is a standard method for preparing alkynes.

(f) Phenol \(\xrightarrow{H_3PO_4} A \xrightarrow{H_2,\) Pt} B \)
Answer:
- A = Benzene (reduction of phenol by heating with zinc dust, though reagent \(H_3PO_4\) or catalytic reduction can lead to dehydration/reduction; typically phenol with Zn dust gives benzene, or catalytic dehydration yields benzene derivatives).
- B = Cyclohexane (\(C_6H_12\)), formed by catalytic hydrogenation of benzene over platinum catalyst (\(H_2 / Pt\)).
In simple words: Reduction of phenol gives benzene, and further hydrogenation of benzene produces cyclohexane.

Exam Tip: Phenol vapors passed over heated zinc dust undergo reduction to form benzene ring.

Additional Reactions:
\(A \xrightarrow{O_3} B \xrightarrow[H_2O / Zn]{[C]} D\)
Answer:
- A = Alkene undergoing ozonolysis.
- B = Ozonide intermediate.
- C / D = Carbonyl compounds (aldehydes or ketones) formed by reductive cleavage of the ozonide with zinc and water.
In simple words: Ozonolysis of alkenes cleaves carbon-carbon double bonds to form carbonyl compounds.

Exam Tip: Zinc dust is used in reductive ozonolysis to prevent further oxidation of aldehydes to carboxylic acids by hydrogen peroxide.